Question

A radioactive material decays by simultaneous emission of two particles with half-lives $1620$ yr and $810$ yr respectively. The time in years after which one-fourth of material remains, is :

• A. $1080$ yr
• B. $2340$ yr
• C. $4860$ yr
• D. $3240$ yr

Answer 1

The correct option is A $1080$ yr

Since, from Rutherford-Soddy law, the number of atoms left after half lives is given by
$N=N_0{\left(\frac{1}{2}\right)}^n$
where, $N_0$ is the original number of atoms.
The number of half lives, $n=\frac{timeofdecay}{effectivehalf-life}$
Relation between effective disintegration constant $\left(\lambda \right)$ and half life $\left(T\right)$
$\lambda =\frac{\ln 2}{T}$
$\therefore {\lambda }_1+{\lambda }_2=\frac{\ln 2}{T_1}+\frac{\ln 2}{T_2}$
Effective half life,
$\frac{1}{T}=\frac{1}{T_1}+\frac{1}{T_2}=\frac{1}{1620}+\frac{1}{810}$
$\frac{1}{T}=\frac{1+2}{1620}\Rightarrow T=540$yr
$\therefore n=\frac{T}{540}$
$\therefore N=N_0{\left(\frac{1}{2}\right)}^{t/540}\Rightarrow \frac{N}{N_0}={\left(\frac{1}{2}\right)}^2={\left(\frac{1}{2}\right)}^{t/540}$
$\Rightarrow \frac{t}{540}=2\Rightarrow t=2\times 540=1080$yr